Question

Chapter P, Section 5, Exercise 134 Find the specified areas for a N ( 0 , 1 ) density. (a) The area below z = 0.8 Round your answer to four decimal places. Area = the absolute tolerance is +/-0.001 (b) The area above z = 1.2 Round your answer to four decimal places. Area = the absolute tolerance is +/-0.001 (c) The area between z = 1.72 and z = 1.26 Round your answer to four decimal places. Area = the absolute tolerance is +/-0.001 Chapter P, Section 5, Exercise 138 Find endpoint(s) on a N ( 0 , 1 ) density with the given property. (a) The area to the left of the endpoint is about 0.68 . Round your answer to two decimal places. Endpoint = the absolute tolerance is +/-0.01 (b) The area to the right of the endpoint is about 0.02 . Round your answer to two decimal places. Endpoint = the absolute tolerance is +/-0.01 (c) The area between ± z is about 0.84 . Round your answer to two decimal places. Endpoint = ± the absolute tolerance is +/-0.01

Answer 1

Chapter P, Section 5, Exercise 134

N(0,1)

(a) The area below z = 0.8

From standard normal tables

P(z < 0.8) = 0.7881

Area = 0.7881

(b) The area above z = 1.2

P( z > 1.2) = 1 - P(z < 1.2)

From standard normal tables

P(z < 1.2) = 0.8849

P( z > 1.2) = 1 - P(z < 1.2) = 1-0.8849 = 0.1151

Area = 0.1151

(c) The area between z = 1.72 and z = 1.26

P(1.26 < z < 1.72) = P(z<1.72) - P(z<1.26)

From standard normal tables,

P(z<1.72) = 0.9573

P(z<1.26) = 0.8962

P(1.26 < z < 1.72) = P(z<1.72) - P(z<1.26) = 0.9573 - 0.8962 = 0.0611

Area = 0.0611

Chapter P, Section 5, Exercise 138

(a) The area to the left of the endpoint is about 0.68

i.e P(z < endpoint) = 0.68

From standard normal tables find the z such that the area against that z is 0.68

end point = 0.47

Endpoint = 0.47

(b) The area to the right of the endpoint is about 0.02

P(z > endpoint) = 0.02

P(z > endpoint) = 1-P(z<endpoint) = 0.02

P(z < endpoint) = 1 - 0.02=0.98

From standard normal tables find the z such that the area against that z is 0.98

endpoint = 2.05

(c) The area between ± z is about 0.84

P(-endpoint < z < endpoint) = 0.84

P(z<endpoint) - p(z < -endpoint) =0.84

because of symmetry,

P(z > endpoint) = P(z<-endpoint)

P(z > endpoint) = 1- P(z<endpoint)

P(z<endpoint) = 1- p(z<-endpoint)

P(z<endpoint) - p(z < -endpoint) =0.84

1 - p(z<-endpoint) - p(z<-endpoint) = 0.4

2P(z<-endpoint) = 1-0.84 = 0.16

p(z<-endpoint) = 0.08

i.e

From standard normal tables find the z such that the area against that z is 0.08

-endpoint = -1.4

Endpoint = 1.4